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Re: 136 theorems on 29 pages
Posted:
Sep 28, 2012 4:44 PM
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Waldek Hebisch schrieb:
>
> clicliclic@freenet.de wrote:
> >
> > Waldek Hebisch schrieb:
> > >
> > > clicliclic@freenet.de wrote:
> > > >
> > > > In my view, the simple, trivial, answer as to why such formulae
> > > > exist is that the "method of undetermined coefficients" works: In
> > > > other words, one may insert polynomial factors and ramp up their
> > > > degree until the linear system of equations that results after
> > > > differentiation becomes soluble.
> > >
> > > Do you have a proof that this works? It seem that if you incresase
> > > degree only on one side, then you end up with multiterm reccurences.
> > > If you increase degree on both sides you have more equations.
> >
> > You are overlooking that there are three sides to the equation: the
> > given integral, the transformed integral, and the integrated term. Extra
> > "numerator" polynomials are inserted into each: the coefficients of that
> > in the given integrand are assumed to be given, those of the other two
> > assumed to be unknown. For a recurrence formula, the polynomials in the
> > two integrands should have the same degree, and the highest admissible
> > degree of that in the integrated term will obviously correlate with
> > that. When the degrees are ramped up simultaneously, the numbers of
> > unknowns rises twice as fast as the number of knowns, and a solution
> > must eventually become feasible. This (fairly obvious) argument could no
> > doubt be formalized into a mathematical proof, but I doubt the author
> > would be interested in this.
>
> What is obvoius is that you do not have a proof -- you deal with
> inhomogeneous system and merely counting number of equations
> is not enough to proof consistency of such system.
>
> And there is quite trivial reason to be suspect of this argument.
> Namely, it does not use assumptions about roots.
I am not particularly interested in a formal proof of this (as opposed
to being interested computer-algebra systems that produce continuous
antiderivatives where possible, and can handle up to and including
elliptic integrands), and the author of the paper doesn't look very
interested either. I agree, of course, that not every linear system of n
equations in n unknowns has solutions. However, for unrestricted
polynomial powers, the roots (and exponents) are unrestricted, and the
formulae in the paper fail precisely in the case of special relations
among the polynomial coefficients (and exponents). This means that the
equations can become dependent, which is no surprise.
Anybody who considers the existence of analogous formulae for higher
degrees in doubt, should certainly try to provide a more formal proof.
He may find the following basic observations helpful: (i) The existence
of single-exponent single-step recurrence formulae implies the existence
of such formulae for any combination of exponent steps (i.e. raising or
lowering of more than one exponent in single or multiple steps). This
follows by iterating the recurrence formulae. (ii) Only relations
involving linear polynomials need be considered, since the "denominator"
polynomials can always be decomposed into linear factors. The piecewise
constant factors caused by branch-cut crossings just drop out of the
relations. (iii) This decomposition automatically takes care of all
degenerate cases, which are turned into lower-order non-degenerate
cases. - What remains to be analyzed, then, are the single-step-up and
single-step-down cases for n linear factors. The minimum degree of the
"numerator" polynomials in the integrands will rise linearly with the
number of linear factors (it turns out to be the number of factors minus
two).
This looks like it could make a nice student exercise. If, on the other
hand, somebody just needs more recurrence relations of this type, or
wants to code a general procedure for friCAS perhaps, he should just
apply the method of undetermined coefficients (and report any failures
please!).
>
> > > >
> > > > [...] Also, there is very little practical interest in going beyond
> > > > elliptic integrals (or beyond Lauricella FD).
> > >
> > > Hmm, that is your opinion. Note however, that when integrating
> > > functions of form R*sqrt(P) where R is a rational function and P is
> > > of degree 3, the rules apply only if R is of degree 1. This seems to
> > > be quite serious limitation. Of course you can split R into partial
> > > fractions, but to get denominator which is power of linear factor may
> > > require extra algebraic extension (in particular may require going to
> > > complex numbers).
> >
> > Let's say, this summarizes my personal experience in science and
> > engineering. I agree with the author that the x^1 and x^3 terms in an
> > elliptic radicand are best annihilated right away to simplify the
> > problem. If the result involves elliptic integrals of the third kind,
> > Pi(x,n,k), the rational cofactor has to be split into partial fractions
> > anyway in order to extract the parameters n, which may therefore come in
> > complex conjugate pairs, etc.
>
> There are lucky cases where reduction process completely solves
> the integral. In other cases, you may need algebraic extensions
> for logarithmic and Pi terms, but algebraic part can be still
> computed without extra extensions.
>
Agreed. But cases where this is absolutely crucial appear to be rare
when problems relating to the physical world are addressed by
integration.
> >
> > One of the pseudo-elliptic examples (here taken from Goursat's Cours
> > d'analyse) is given on page 50 of the booklet mentioned.
>
> AFAICS this example is wrong -- FriCAS result indicates that
> there is no elementary integral in this case and I see no
> reasons why it should have one. OTOH, if one modifies
> the example to have
>
> f/sqrt((1 - x^2)*(1 - k^4*x^2))
>
> under integral, then FriCAS can handle it for particular f-s.
>
Thanks for trying. My Gradshteyn-Ryzhik has three such integrands and
adds more detail (without specifying the source). In particular, the
substitution z*x = SQRT(x*(1-x)*(1-k^2*x)) is said to turn the integrand
into a rational function. I will look into this claim. The other two
examples are the same integrand f(x)/sqrt(x*(1-x)*(1-k^2*x)) with f(x) +
f((1-k^2*x)/(k^2*(1-x))) = 0 and with f(x) + f((1-x)/(1-k^2*x)) = 0.
> > >
> > > Well, there is more conditions beyond confluent roots, so just
> > > excluding them is not enough -- the user still have to check for
> > > division by zero. However, the point of my remark was that repeated
> > > blurb about confluent roots is more distraction than help for the
> > > reader. Of course, the author is free to ignore my opinion...
> >
> > I expect your opinion will be appreciated by the author. As far as I can
> > see, excluding additional confluences at any level of degeneracy is a
> > sufficient condition (though not a necessary condition) to prevent zero
> > division: Some confluences play no role (e.g. 4*d*f = e^2 doesn't affect
> > 8.1.1, 8.1.2, 8.1.3 in the paper at all) whereas others play a selective
> > role (e.g. 4*a*c = b^2 affects 8.1.1 and 8.1.3 in the paper but doesn't
> > affect 8.1.2). While simply excluding confluent roots then seems to be
> > enough indeed, it also does no harm since those integrands are dealt
> > with separately anyway.
>
> AFAICS there are three kinds of conditions:
>
> 1) Nonzero leading terms (assumed to hold globally)
> 2) confluent roots
> 3) exceptional values of exponents, for example 3.3.2 contains
> factor (n + 1)*(2*n + 1), so -1 and -1/2 must be excluded.
>
> In the intended range, that is increasing exponents toward
> -1 or decreasing them toward 0 all exponents are good. But
> outside of that range there are exceptional values.
>
I see. It looked like you were concerned about some problem with the
discriminants/resultants, which can be addressed by excluding
(additional) confluent roots. Nonzero leading coefficients imply that
higher-order formulae cannot be used for lower-order problems right
away. Still, lower-order formulae can be recovered from higher-order
ones by taking suitable limits (the world expert in this is Albert
Rich). If one hits upon a factor like (m+n+1)*(m+n+2) in formula (8.1.2)
of the paper, the obstacle can be moved out of the way by switching to a
recurrence relation for the other exponent (see Rubi). Warning: Again, I
haven't analyzed this issue formally to arbitray order, I simply find no
reason to doubt this conclusion based on experience.
Martin.
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